2015 Mathematics Higher Finalised Marking Instructions · Page 2 General Comments These marking...

2015 Mathematics

Higher

Finalised Marking Instructions

Scottish Qualifications Authority 2015 The information in this publication may be reproduced to support SQA qualifications only on a non-commercial basis. If it is to be used for any other purposes written permission must be obtained from SQA’s NQ Delivery: Exam Operations team. Where the publication includes materials from sources other than SQA (secondary copyright), this material should only be reproduced for the purposes of examination or assessment. If it needs to be reproduced for any other purpose it is the centre’s responsibility to obtain the necessary copyright clearance. SQA’s NQ Delivery: Exam Operations team may be able to direct you to the secondary sources. These Marking Instructions have been prepared by Examination Teams for use by SQA Appointed Markers when marking External Course Assessments. This publication must not be reproduced for commercial or trade purposes.

General Comments

These marking instructions are for use with the 2015 Higher Mathematics Examination. For each question the marking instructions are in two sections, namely Illustrative Scheme and Generic Scheme. The Illustrative Scheme covers methods which are commonly seen throughout the marking. The Generic Scheme indicates the rationale for which each mark is awarded. In general, markers should use the Illustrative Scheme and only use the Generic Scheme where a candidate has used a method not covered in the Illustrative Scheme. All markers should apply the following general marking principles throughout their marking: 1 Marks must be assigned in accordance with these marking instructions. In principle,

marks are awarded for what is correct, rather than deducted for what is wrong. 2 One mark is available for each . There are no half marks. 3 Working subsequent to an error must be followed through, with possible full marks for

the subsequent working, provided that the level of difficulty involved is approximately similar. Where, subsequent to an error, the working for a follow through mark has been eased, the follow through mark cannot be awarded.

4 As indicated on the front of the question paper, full credit should only be given where

the solution contains appropriate working. Throughout this paper, unless specifically mentioned in the marking instructions, a correct answer with no working receives no credit.

5 In general, as a consequence of an error perceived to be trivial, casual or insignificant,

e.g. candidates lose the opportunity of gaining a mark. But note the second

example in comment 7. 6 Where a transcription error (paper to script or within script) occurs, the candidate

should be penalised, e.g.

2 5 7 9 4

4 3 0

( 3)( 1) 0

1 or 3

x x x

x x

x x

x

This is a transcription error and so the mark is not awarded.

Eased as no longer a solution of a quadratic equation.

Exceptionally this error is not treated as a transcription error as the candidate deals with the intended quadratic equation. The candidate has been given the benefit of the doubt.

~~

~

7 Vertical/horizontal marking Where a question results in two pairs of solutions, this technique should be applied, but

only if indicated in the detailed marking instructions for the question. Example: Point of intersection of line with curve

Illustrative Scheme: 5 2 5x y

6 4 7x y

Markers should choose whichever method benefits the candidate, but not a combination of both.

8 In final answers, numerical values should be simplified as far as possible, unless

specifically mentioned in the detailed marking instructions.

Examples: should be simplified to or should be simplified to 43

should be simplified to 50 should be simplified to

must be simplified to 8

9 Commonly Observed Responses (COR) are shown in the marking instructions to help mark common and/or non-routine solutions. CORs may also be used as a guide when marking similar non-routine candidate responses.

10 Unless specifically mentioned in the marking instructions, the following should not be

penalised: Working subsequent to a correct answer;

Correct working in the wrong part of a question;

Legitimate variations in numerical answers, e.g. angles in degrees rounded to nearest degree;

Omission of units;

Bad form (bad form only becomes bad form if subsequent working is correct), e.g.

3 2

3 2

4 3 2 3 2

4 3 2

( 2 3 2)(2 1)

written as

( 2 3 2) 2 1

2 4 6 4 2 3 2

2 5 8 7 2 gains full credit;

x x x x

x x x x

x x x x x x x

x x x x

Repeated error within a question, but not between questions.

11 In any ‘Show that . . .’ question, where the candidate has to arrive at a required result, the last mark of that part is not available as a follow through from a previous error unless specifically stated otherwise in the detailed marking instructions.

12 All working should be carefully checked, even where a fundamental misunderstanding is

apparent early in the candidate’s response. Marks may still be available later in the question so reference must be made continually to the marking instructions.

The square root of perfect squares up to and including 100 must be known.

5

6

All working must be checked: the appearance of the correct answer does not necessarily indicate that the candidate has gained all the available marks.

13 If you are in serious doubt whether a mark should or should not be awarded, consult

your Team Leader (TL). 14 Scored out working which has not been replaced should be marked where still legible.

However, if the scored out working has been replaced, only the work which has not been scored out should be marked.

15 Where a candidate has made multiple attempts using the same strategy, mark all

attempts and award the lowest mark. Where a candidate has tried different strategies, apply the above ruling to attempts within each strategy and then award the highest resultant mark. For example:

Strategy 1 attempt 1 is worth 3 marks Strategy 2 attempt 1 is worth 1 mark

Strategy 1 attempt 2 is worth 4 marks Strategy 2 attempt 2 is worth 5 marks

From the attempts using strategy 1, the resultant mark would be 3.

From the attempts using strategy 2, the resultant mark would be 1.

In this case, award 3 marks.

16 In cases of difficulty, covered neither in detail nor in principle in these instructions,

markers should contact their TL in the first instance.

Paper 1 Section A

Question Answer 1 C

2 B

3 D

4 A

5 C

6 B

7 C

8 D

9 B

10 D

11 C

12 C

13 B

14 D

15 A

16 B

17 D

18 C

19 B

20 A Summary A 3

B 6

C 6

D 5

Question Generic Scheme Illustrative Scheme Max Mark

21(a).

1 know to use 1x

2 interpret result and state conclusion

3 state quadratic factor

4 factorise completely

Method 1

1 3 21 6(1) 9(1) 4

2 0 ( 1) is a factor.x

Method 2

1

1 1 6 9 4

1

1

2

1 1 6 9 4

1 5 4

1 5 4 0

remainder 0 ( 1) is a factor.x

Method 3

1

2

3 2

3 2

1 6 9 4

x

x x x x

x x

2 0 ( 1) is a factor.x

3 2 5 4x x stated or implied

by 4

4 2

1 4x x 4

Paper 1 - Section B

Notes:

1. Communication at 2 must be consistent with working at that stage i.e. a candidate’s

working must arrive legitimately at 0 before 2 is awarded.

2. Accept any of the following for 2:

‘ (1) 0 so ( 1)is a factorf x ’

‘since remainder is 0 , it is a factor’

the 0 from the table linked to the word ‘factor’ by e.g. ‘so’, ‘hence’, ‘ ’, ‘ ’,

‘ ’

3. Do not accept any of the following for 2:

double underlining the zero or boxing the zero without comment

‘ 1 is a factorx ’, ‘ ( 1) is a factorx ’, ‘ 1 is a rootx ’, ‘ ( 1) is a rootx ’, “ ( 1)x is

a root”

the word ‘factor’ only, with no link

4. At 4 the expression may be written as 1 1 4x x x in any order.

5. An incorrect quadratic correctly factorised may gain 4. 6. Where the quadratic factor obtained is irreducible, candidates must clearly demonstrate

that 2 4 0b ac to gain 4.

7. 0 must appear at 1 or

2 for 2 to be awarded.

8. For candidates who do not arrive at 0 at the 2 stage 2

3

4 are not available. 9. Do not penalise candidates who attempt to solve a cubic equation. However, within this working there may be evidence of the correct factorisation of the cubic.

10. Evidence for 3&

4 may appear in part (b).

Commonly Observed Responses:


21(b)(i).

5 know to and differentiate

6 find gradient

7 state equation of tangent

5 23 12 11x x

6 2

7 2 1y x 3

Notes:

11. 7 is only available if an attempt has been made to find the gradient from

differentiation.

12. At mark 7 accept 3 2( 1)y x , 2 1y x or any other rearrangement of the

equation.


Candidate A

5

6 using y mx c

1 3 2

3 2 1

1

2 1

x y m

c

c

y x

21(b)(ii).

8 set CURVE LINEy y

9 arrange equation in standard cubic form

10 identify x coordinate of B and calculate y

coordinate

8 3 26 11 3 2 1x x x x

9 3 26 9 4 0x x x

10(4,9) 3

Notes:

13. 9 is only available if ‘ 0 ’ appears in at least one arrangement of the equation.

14. Solutions at 10 must be consistent with working at 4 and 7.

15. Candidates who obtain three distinct factors at 4 can gain 8 and 9 but 10 is unavailable.

16. For 10 accept 4, 9x y .

17. Do not penalise the appearance of (1,3) .


7


22.

1 arrange in differentiable form

2 start differentiation

3 complete differentiation and set ( ) 0f x

4 evaluate f at stationary point

5 consider end-points

6 state max and min values

1 2( ) 4f x x x

2 38 or 1x

3 38 1 0x

4 2, ( ) 3x f x

5 17

(1) 5, (4) (or 4 25)4

f f

6 max5, min3 6

Notes:

1. Candidates must attempt to differentiate a term with a –ve or fractional power by 3 for 3 to be awarded.

2. 3 is not available for simply stating ‘ '( ) 0f x ’. A clear link between the candidates

derivative and ‘ '( ) 0f x ’is required.

3. For candidates who integrate, but clearly believe they are finding the derivative 1 4 5 6 are available (see CORs – K, L, and M). In other instances where candidates have integrated

then only 1 and 5 are available. A numerical approach can only gain 5.

4. 5 and 6 are not available to candidates who consider stationary points only.

5. Treat maximum (1,5) and minimum (2,3) as bad form.

6. Vertical marking is not applicable to 5 and 6.

7. The appearance of (2,3) following any 2nd derivative or nature table gains 4.

8. If at the 4 stage a value of x is obtained outwith the given interval then 4 is unavailable,

but 6 may still be gained (see CORs – F and G). 9. Candidates who consider the end values but do not evaluate the stationary value cannot

gain 6.


Candidate A

2

3

( ) 4

'( ) 8 1

for stationary

points '( ) 0

f x x x

f x x

f x

With no further working

Candidate B

2( ) 4

for stationary

points '( ) 0

f x x x

f x


Candidate C

2

3

( ) 4

'( ) 8 1

for stationary

points 0

f x x x

f x x

dy

dx


Candidate D

2

3

( ) 4

8 1

for stationary

points '( ) 0

f x x x

x

f x


1

2

3

1

2 ^

3 ^

1

2

3 ^

1

2

3 ^

Candidate E Solely a numerical attempt

(1) 5, (2) 3,

31 17(3) , (4)

9 4

f f

f f

Award only 5

For any similar attempt which includes the

evaluation of f for a value

outwith the range award 0.

Candidate F

2( ) 4

'( ) 8 ...........

'( ) ........... 1 0

1 1 7( ) 255

8 8 8

1(1) 5, (4) 4

4

1 7min 4 , max 255

4 8

f x x x

f x x

f x

x f

f f

Candidate G

2( ) 4

'( ) 8 ...........

'( ) ........... 1 0

1 1 1( ) 256

8 8 8

1(1) 5, (4) 4

4

1 1min 4 , max 256

4 8

f x x x

f x x

f x

x f

f f

Candidate H

2

1

( ) 4

'( ) 8 ...........

'( ) ........... 1 0

18 ( ) 8

16

1(1) 5, (4) 4

4

1 1min 4 , max 8

4 16

f x x x

f x x

f x

x f x

f f

Candidate I

3

2

( ) 4

'( ) 0 3 0

0 ( ) undefined

f x x

f x x

x f x

Candidate J

2

4'( ) 0

1(1) 5, (4) 4

4

f x xx

f f

Candidate K

2

21

( ) 4

'( ) 42

2 (2) 3

1(1) 5, (4) 4

4

min 3, max 5

f x x x

xf x x

x f

f f

Candidate L

2

22 1

( ) 4

4 4 02

2 (2) 3

1(1) 5, (4) 4

4

min 3, max 5

f x x x

xx x x

x f

f f

Candidate M

2

21

4

4 02

2 (2) 3

1(1) 5, (4) 4

4

min 3, max 5

x x

xx

x f

f f

1

2 ×

3

4

5

6

1

2 ×

3

4

5

6

1

×

2 ×

3 ×

4 ×

5

6 ^

1

×

2 ×

3

4 ×

5 ^

6 ^

1 1

1

2 ×

3

4

5

6

1

1

2 ×

3 ×

4

5

6

1

1

1

2 ×

3 ×

4 ×

5

6 ×

1

2 ×

3 ×

4 ×

5

6 ×


23.

1 collect log terms

2 use laws of logs

3 use laws of logs

4 solve for x

Method 1

1 2 2log 3 7 log 1 3x x

stated or implied by 2

2

2

3 7log 3

1

x

x

3

33 7

21

x

x

4 3x

Method 2

1

2 2 2log 3 7 3log 2 log 1x x

stated or implied by 2

2

3

2 2 2log 3 7 log 2 log ( 1)x x

3 2 2log 3 7 log 8( 1)x x

4 3x 4

Notes:

1. For 3 accept

2 2

3 7log log 8

1

x

x

.


Candidate A

2 2

2 2

2

3

log 3 7 3 log 1

3 log 1 log 3 7

13 log

3 7

12

3 7

8 8 3 7

3

x x

x x

x

x

x

x

x x

x

Candidate B

2 2

2 2

2

3

2

log (3 7) 3 log ( 1)

log (3 7) log ( 1) 3

log (3 7)( 1) 3

(3 7)( 1) 2

3 4 15 0

(3 5)( 3) 0

5or 3

3

discard as 1

5

3

x x

x x

x x

x x

x x

x x

x x

x

x

1

2

3

4

1

×

2

3

4

4 is not available

for candidates who

do not discard

3x


24.

1 interpret the values of a, b and c and substitute

2 know to use discriminant 0

Method 1

3 simplify or factorise quadratic inequation

4 state range of values of k

Method 2

3 simplify or factorise quadratic expression

4 evidence and range of values of k

1 23 4 9k k

2 .... 0

Method 1

3

2 9or 9 1 2 1 2 0

36k k k

4 1 1

2 2k

Method 2

3 2 1 19 36 0 ,

2 2k k

4 graph or other evidence

leading to 1 1

2 2k

4

Notes:

1. The “ 0 ”must appear at least once at the 1 or 2 stage for 2 to be awarded.

2. If an x appears in the candidate’s ‘discriminant’ only 2 may be awarded.

3. The use of any expression masquerading as the discriminant can gain only 2 at most.

4. Award 2 to candidates who write BOTH 29 36 0k AND 29 36 0k .

5. For candidates who at 3 simplify or factorise an equation 4 can only be awarded if evidence of solving an inequation (for example a graph) appears.

6. At 2 stage, quoting 2 4 0b ac is not sufficient.

7. At 3 stage, in Method 2, solutions for k need not be simplified.



25(a)

1 know to use Theorem of Pythagoras’ or distance formula

2 process to obtain result

1 2 2

2 5 10D t t

2 2 2

2

4 20 25 20 100

5 40 125

D t t t t

t t

2

Notes:

Be wary of fudged solutions!


Candidate A

2 22

2 1 2 1

2 2

2

(2 5) 0 0 ( 10)

5 40 125

D y y x x

D t t

D t t

See note 12 in the general instructions

Beware

2 2

2 2

2

(2 5,0) (0, 10)

0 2 5 2 5

10 0 10

(2 5) ( 10)

4 20 20 25 100

5 40 125

A t B t

t tAB

t t

D t t

D t t t t

D t t

NB: This is an exception to note 12 in the general instructions

25(b)

3 write in differentiable form

4 start differentiation

5 complete differentiation

6 substitute t = 5 and interpret result

3 1

2 25 40 125t t

4 1

2 21

5 40 125 .....2

t t

5 ..... 10 40t

6 10

(5) 0 increasing2 50

D 4

Notes:

1. 4 is only available for differentiating an expression of the form (trinomial)proper fraction.

2. Do not penalise the use of x instead of t .

3. 6 is only available to candidates who substitute into a derivative.

1

×

2 ×

1

×

2

1


Candidate A

12

22

2

( ) 5 40 125

( ) 5 40 125 ...

..... (10 40)

10(5) 0 decreasing

50

D t t t

D t t t

t

D

Candidate B

12 2

32 2

3

( ) 5 40 125

1( ) 5 40 125 ...

2

..... (10 40)

5(5) 0 decreasing

50

D t t t

D t t t

t

D

Candidate C

12 2

32 2

3

( ) 5 40 125

1( ) 5 40 125 ...

2

..... (10 40)

(5) 5 50 0 increasing

D t t t

D t t t

t

D

Candidate D

12 2

12 2

12 2

( ) 5 40 125

1( ) 5 40 125 10 40

2

( ) 5 5 40 125 40

25(5) 40 0 decreasing

50

D t t t

D t t t t

D t t t t

D

Candidate E

12 2

12 2

( ) 5 40 125

1( ) 5 40 125 10 40

2

1(5) 0 increasing

2

D t t t

D t t t t

D

Candidate F – Alternative Method

2 2

2

2

22

5 40 125

10 40

5 50 40

10 0 is increasing

D is increasing

D t t

dDt

dt

dDt

dt

dDD

dt

Calculating Distance

5 50

4 45

so distance is increasing

t D

t D

Award 0 marks as answer is not from differentiation.

Alternative Response

3

×

4 × see note 1.

5

6

1

3

×

4

5

6

1

1

3

4 ×

5

6

1

3

4

5 ×

6

1

3

4

5

6

3

4

5

6

2 2

2 2

5( 8 25)

5[( 4) 9]

D t t

D t

Graph together with a statement indicating that

when 25,t D is increasing

and therefore D is increasing.

Min TP at 4x

3

4 ×

5 ×

6 ×

3

4

5

6

Paper 2


1(a)

1 calculate gradient of AB

2 use property of perpendicular lines

3 substitute into general equation of a line

4 demonstrate result

1 3ABm

2 1

3altm

3 1

3 133

y x

4 ... 3 4x y 4

Notes:

1. 3 is only available as a consequence of trying to find and use a perpendicular gradient.

2. 4 is only available if there is/are appropriate intermediate lines of working between 3

and

4. 3. The ONLY acceptable variations for the final equation for the line in 4 are

4 3 , 3 4, 4 3x y y x y x .


Candidate A

AB

1 5 4 1

5 7 12 3

3

3 3 13

alt

m

m

y x

4 is not available

Candidate B

For 4

1 133

3 3

1 4

3 3

3 4 -

3 4 -

3 4

y x

y x

y x

y x

x y

not acceptable

not acceptable

1

1

1 ×

2

3

4 ×


1(b)

5 calculate midpoint of AC

6 calculate gradient of median

7 determine equation of median

5 ACM (4,5)

6 2BMm

7 2 3y x 3

Notes:

4. 6 and 7 are not available to candidates who do not use a midpoint. 5. 7 is only available as a consequence of using a non-perpendicular gradient and a midpoint. 6. Candidates who find either the median through A or the median through C or a side of the triangle gain 1 mark out of 3.

7. At 7 accept ( 5) 2( ( 1))y x , 5 2( 4)y x , 2 3 0y x or any other rearrangement of

the equation.


Median through A

M (6, 1)

8

11

8 81 ( 6) or 7 ( 5)

11 11

BC

AMm

y x y x

Award 1/3

Median through C

M ( 3,1)

1

8

1 13 ( 13) or 1 ( 3)

8 8

AB

CMm

y x y x

Award 1/3

1(c)

8 calculate x or y coordinate

9 calculate remaining coordinate of the point of intersection

8 1 or 1x y

9 1 or 1y x 2

Notes:

8. If the candidate’s ‘median’ is either a vertical or horizontal line then award 1 out of 2 if both coordinates are correct, otherwise award 0.


For candidates who find the altitude through B in part (b)

1

5x

7

5y

Candidate A

(b)5 2( 4)

2 13 -error

y x

y x

(c)3 4

2 13

x y

y x

Leading to 7 and 1x y

8

9

1

1

7

8 ×

9

1


2(a)

1 interpret notation

2 state a correct expression

1 ((1 )(3 ) 2)f x x stated or

implied by 2

2 10 (1 )(3 ) 2x x stated or

implied by 3 2

Notes:

1. 1 is not available for (10 )g f x g x but 2 may be awarded for 1 10 3 (10 ) 2x x .


Candidate A

(a)( ( )) ' ( ( )) '

(1 10 )(3 (10 )) 2

f g x g f x

x x

2 2

2

2

2

75 18 or 18 75

( 18

( 9)

( 9) 6

x x x x

x x

x

x

2( 9) 6 0

9 6 or 9 6

x

x

Candidate B

( ( ))

10((1 ) (3 )) 2

f g x

x x

Candidate C

( ( ))

10((1 )(3 ) 2)

f g x

x x

2(b)

3 write ( ( ))f g x in quadratic form

Method 1

4 identify common factor

5 complete the square

Method 2

4 expand completed square form and equate coefficients

5 process for andq r and write in required

form

3 2 215 2 or 2 15x x x x

Method 1

4 21( 2x x stated or implied

by 5

5 21( 1) 16x

Method 2

4 2 22 and 1,px pqx pq r p

5 1 and 16q r

Note if 1p 5 is not available

3

1

×

2

3

4

5

6

7

1

1

1

1

1

1 (b)

(c)

1

^

2 ×

1

^

2 ×

Notes:

2. Accept 216 ( 1)x or 2( 1) 16x at 5.


Candidate A

2

2

2

( 2 15)

( 2 1 1 15)

( 1) 16

x x

x x

x

Candidate B 2

2

2 2

15 2

2 15

2 and 1

1 16

x x

x x

px pqx pq r p

q r

Candidate C

2

2

2

2 15

( 1) ...

( 1) 14

x x

x

x

Candidate D

2

2

2

15 2

2 15

( 1) 16

x x

x x

x

Eased, unitary coefficient of 2x

(lower level skill)

Candidate E

2

2

2

2 2

15 2

2 15

( 1) 16

so 15 2 ( 1) 16

x x

x x

x

x x x

Candidate F

2

2

2

2 15

( 1) ...

( 1) 16

x x

x

x

2(c)

6 identify critical condition

7 identify critical values

6 21( 1) 16 0x

or (( ( )) 0f g x

7 5 and 3 2

Notes:

3. Any communication indicating that the denominator cannot be zero gains 6.

4. Accept 5 and 3 or 5 and 3x x x x at 7.

5. If 5 and 3x x appear without working award 1/2.


Candidate A

2

1

( 1) 16

5

x

x

Candidate B

1

( ( ))

( ( )) 0

3, 5

3 5

f g x

f g x

x x

x x

3(a)

1 determine the value of the required term 1

3 9122 or or 22 75

4 4

1

Notes:

1. Do not penalise the inclusion of incorrect units. 2. Accept rounded and unsimplified answers following evidence of correct substitution.


3

4 ×

5

3

4

eased

6

7 ^

6 ×

7

4

5 ×

3

4 ×

5

eased

3

4 ×

5 ×

3

4 ×

5

1

5


3(b)

Method 1 (Considering both limits)

2 know how to calculate limit


4 calculate limit

5 calculate limit

6 interpret limits and state conclusion

Method 2

(Frog first then numerical for toad)


3 calculate limit

4 determine the value of the highest term less than 50

5 determine the value of the lowest term greater than 50

6 interpret information and state conclusion

Method 3 (Numerical method for toad only)

2 continues numerical strategy

3 exact value

4 determine the value of the highest term less than 50

5 determine the value of the lowest term greater than 50

6 interpret information and state conclusion

Method 4 (Limit method for toad only)

2 & 3 know how to calculate limit

4 &

5 calculate limit

6 interpret limit and state conclusion

Method 1

2 13

32

1 or

1L L 32

3

3 34

13

1 or

3L L 13

4

4 48

5 52

6 52 50 toad will escape

Method 2

2 13

32

1or

1L L 32

3

3 48

4 49 803...

5 50 352...

6 50 352 50 toad will escape

Method 3

2 numerical strategy

3 30 0625

4 49 803...

5 50 352...

6 50 352 50 toad will escape

Method 4

2 & 3

34

13

1 or

3L L 13

4

4 &

5 52

6 52 50 toad will escape 5

Notes:

3. 6 is unavailable for candidates who do not consider the toad in their conclusion.

4. For candidates who only consider the frog numerically award 1/5 for the strategy.


Error with frogs limit – Frog Only

F

F

34L

11

3

L 51

51 50

frog will escape.

Using Method 3 - Toad Only

Using Method 3- Toad Only

Using Method 3 - Toad Only

Toad Conclusions Limit = 52

This is greater than the height of the well and so the toad will escape – award 6.

However

Limit =52 and so the toad escapes - 6 ^.

Iterations

1

2

3

4

5

6

7

8

9

32

42 667

46 222

47 407

47 802

47 934

47 978

47 993

47 998

f

f

f

f

f

f

f

f

f

1

2

3

4

5

6

7

8

9

10

11

12

13

22 75

30 0625

35 547

39 660

42 745

45 059

46 794

48 096

49 072

49 804

50 353

t

t

t

t

t

t

t

t

t

t

t

t

2 ×

3 ×

4

5

6

1

1

1

2

3

4 missing ^

5 50 352...

6 50.352 50

so the toad escapes.

2

3

4 missing ^

5 50 1.. rounding error ×

6 50.1 50


2

3

4 49 7.. rounding

error ×

5 50 1...

6 50.1 50


1

1

1


4(a)

1 know to equate ( ) and ( )f x g x

2 solve for x 1

2 21 1 1 33 5

4 2 4 2x x x x

2 2x 2

Notes:

1. 1 and 2 are not available to candidates who: (i) equate zeros, (ii) give answer only

without working, (iii) arrive at 2x with erroneous working.


Candidate A

2

2

1 13

4 2

1 35

4 2

subtract to get

0 2

2

y x x

y x x

x

x

Candidate B

2

2

1 13

4 2

1 35

4 2

2

x x

x x

x

In this case the candidate has equated zeros

Candidate C

2 21 1 1 3( ) 3 ( ) 5

4 2 4 2

1 1 1 3'( ) '( )

2 2 2 2

. .

1 3

2

f x x x g x x x

f x x g x x

x x

x

1

2

1

×

2 ×

1

2


4(b)

3 know to integrate

4 interpret limits

5 use ‘upper – lower’

6 integrate

7 substitute limits

8 evaluate area between ( ) and ( )f x h x

9 state total area

3

4 2

0

5

2

2 23 91 14 2 8 4

0

( 3) ( 3)x x x x dx

6 3 21 724 8x x accept

unsimplified integral

7 3 21 724 8

2 2 0

8 196

9 193

7

Notes:

2. If limits 0 and 2x x appear ex nihilo award 4.

4. If a candidate differentiates at 6 then 6, 7 and 8 are not available. However, 9 is still available.

5. Candidates who substitute at 7, without attempting to integrate at 6, cannot gain 6, 7 or

8. However, 9 is still available.

6. Evidence for 8 may be implied by 9.

7. 9 is a strategy mark and should be awarded for correctly multiplying their solution at 8, or

for any other valid strategy applied to previous working.

8. For 5 both a term containing a variable and the constant term must be dealt with correctly.

9. In cases where 5 is not awarded, 6 may be gained for integrating an expression of

equivalent difficulty i.e. a polynomial of at least degree two. 6 is unavailable for the integration of a linear expression.

10. 8 must be as a consequence of substituting into a term where the power of x is not equalto 1 or 0.


Candidate A – Valid Strategy

Candidate B – Invalid Strategy For example, candidates who integrate each of the four functions separately within an invalid strategy

Candidate C 2

2 23 91 14 2 8 4

0

2

21 118 4

0

3 2

( 3 3)

( )

1 11

24 8

x x x x dx

x x dx

x x

Candidate D 2

2 23 91 14 2 8 4

0

2

21 118 4

0

3 2

( 3 3)

( 6)

1 116

24 8

x x x x dx

x x dx

x x x

Candidate E

1 1... cannot be negative so

3 3

1 1however, so Area

3 3

Candidate F 2

2 23 91 14 2 8 4

0

2

218

0

3 27124 8

( 3 3)

7( )

4

x x x x dx

x x dx

x x

A B

Candidates who use the strategy:

Total Area = Area A + Area B Then mark as follows:

Mark Area A for 3 to

8 then

mark Area B for 3 to

8 and

award the higher of the two.

9 is available for correctly

adding two equal areas.

3

Gain 4 if limits correct on

( ) and ( )

or

( ) and ( )

f x h x

g x k x

5 is unavailable

Gain 6 for calculating either

( ) or ( )

and

( ) or ( )

f x g x

h x k x

Gain 7 for correctly substituting at least twice

Gain 8 for evaluating at least two integrals

correctly

9 is unavailable

8

×

8

~~~~~~~~~

5

6

×

5

×

6

1

~~~~

5

6


5(a)

1 state centre of C1

2 state radius of C1

3 calculate distance between centres of C1 and

C2

4 calculate radius of C2

1 ( 3, 5)

2 5

3 20

4 15 4

Notes:

1. For 4 to be awarded radius of C2 must be greater than the radius of C1. 2. Beware of candidates who arrive at the correct solution by finding the point of contact

by an invalid strategy.

3. 4 is for c1c2 c1Distance r but only if the answer obtained is greater than c1r .



5(b)

5 find ratio in which centre of C3 divides line joining centres of C1 and C2

6 determine centre of C3

7 calculate radius of C3

8 state equation of C3

5 3 : 1

6 (6,7)

7 20r (answer must be consistent with distance between centres)

8 2 2( 6) ( 7) 400x y 4

Notes:

4. For 5 accept ratios 3: 1, 1: 3, 3: 1, 1: 3 (or the appearance of 3

4).

5. 7 is for c2 c1r +r .

6. Where candidates arrive at an incorrect centre or radius from working then 8 is available.

However 8 is not available if either centre or radius appear ex nihilo (see note 5).

7. Do not accept 220 for 8.

8. For candidates finding the centre by ‘stepping out’ the following is the minimum evidence

for 5 and 6:


Candidate A

3

3

2 2

using the mid-point of centres:

centre C (3,3)

radius of C 20

( 3) ( 3) 400x y

Candidate B

1 2

3

3

2 2

C ( 3, 5) C (9,11) r 20

1: 3

01

44

(0, 1)

( 1) 400

C

C

x y

Candidate C – touches 1C internally only

Candidate D - touches 2C internally only

Candidate E – centre 3C collinear with 1 2C ,C

5

×

6

7

8

1

5 note 4

6

7

8

5 ×

6 3centre C (3,3) ×

7 3 2radius of C radius of C 15

8

2 2( 3) ( 3) 225x y

1

1

5 ×

6 3centre C (3,3) ×

7 3 1radius of C radius of C 5

8

2 2( 3) ( 3) 25x y

1

1

5 ×

6 3e.g. centre C (21,27) ×

7 3 1radius of C 45 (touch C internally only)

8

2 2( 21) ( 27) 2025x y

1

1

1

3 4

Correct ‘follow through’

using the ratio 1:3

16

12 12

9

16

Correct answer using

the ratio 3:1

12

5

6

×

5

6


6(a)

1 Expands

2 Evaluate p.q

3 Completes evaluation

1

p.q p.r

2 1

42

3

1... 0 4

2

3

Notes:

1. For .( ) p q r pq pr with no other working 1 is not available.


6(b)

4 correct expression 4 - or equivalentq p r 1

6(c)

5 correct substitution

6 start evaluation

7 find expression for r

5 - q.q q.p q.r

6 o 9

9 .... 3 cos30 9 32

r

7

3 3

cos30r

3

Notes:

2. Award 5 for 2- q +q.p q.r


Candidate A

o

9- 9 3

2

9 99 3 cos150 9 3

2 2

3 3

cos150

q.q q.p q.r =

r

r

Candidate B

o

9- 9 3

2

9 99 3 cos30 9 3

2 2

6

q.q q.p q.r =

r

r

5

6

×

7

5

6

7

1


7(a)

1 integrate a term

2 complete integration with constant

1

3sin 2

2x x

2

3sin 2

2x c x c

2

Notes:


7(b)

3 substitute for cos 2x

4 substitute for 1 and complete

3

2 2

2 2

3 cos sin ...

or ...(sin cos )

x x

x x

4

2 2 2 2... sin cos 4cos 2sinx x x x

2

Notes:

1. Any valid substitution for cos2x is acceptable for 3.

2. Candidates who show that 2 24cos 2sin 3cos2 1x x x may gain both marks.

3. Candidates who quote the formula for cos2x in terms of A but do not use in the context of

the question cannot gain 3.


Candidate A 2 2 2

2 2

3cos 2 1 (2cos 1) (2cos 1) (1 2sin ) 1

4cos 2sin

x x x x

x x

Candidate B 2 24cos 2sin 2(cos 2 1) (1 cos 2 )

3cos 2 1

x x x x

x

7(c)

5 interpret link

6 state result

5

1...

2

6

3 1sin 2

4 2x x c

2

Notes:


Candidate A 2 2sin 2cos

(3cos 2 1)

3sin 2

2

x x dx

x dx

x x c

3

4

3

4

5

×

6

×

OR


8.

1 use compound angle formula

2 compare coefficients

3 process for k

4 process for a

5 equates expression for h to 100

6 write in standard format and attempt to

solve

7 solve equation for 1 5t

8 process solutions for t

1 sin1 5 cos cos1 5 sink t a k t a

2

k cos a = 36, k sin a = 15 stated

explicitly

3

k = 39

4 0·39479 rad or 22 6a

5

39sin 1 5 0 39479... 65 100t

6

35sin 1 5 0 39479...

39t

1 35

1 5 0 39479... sin39

t

7

8

7

1 5 1 508t

and 2·422

8 t = 1·006 and 1·615

8

Notes:

1. Treat sin cos cos si1 1 5 n5k t a t a as bad form only if the equations at the 2 stage both

contain k.

2. 39sin cos 39co1 5 1 5s sint a t a or 39 sin cos cos si1 5 1 5 nt a t a is acceptable for 1 and

3.

3. Accept cos 36 and sin 15k a k a for 2.

4. 2 is not available for cos 36 and sin1 5 1 155k t k t , however, 4 is still available.

5. 3 is only available for a single value of , 0k k .

6. 4 is only available for a single value of a .

7. The angle at 4 must be consistent with the equations at 2 even when this leads to an angle outwith the required range.

8. Candidates who identify and use any form of the wave equation may gain 1 , 2 and 3 ,

however, 4 is only available if the value of a is interpreted for the form sin(1 5 )k t a .

9. Candidates who work consistently in degrees cannot gain 8.

10. Do not penalise additional solutions at 8.

11. On this occasion accept any answers which round to 1 0 and 1 6 (2 significant figures

required).


Response 1: Missing information in working.

Candidate A

39cos 36

39sin 15

15tan

36

0·39479 rad or 22 6

a

a

a

a

Candidate B cos 36

sin 1 5

15tan

36

0·39479 rad or 22 6

a

a

a

a

Candidate C

sin1 5 cos cos1 5 sin

cos 36, sin 1 5

39 or 39

15tan

36

0·39479 rad or 22 6

3 53638...rad or 202 6

k t a k t a

k a k a

k

a

a

or

a

Response 2: Correct expansion of k sin( x + a)º and possible errors for 2 and 4

Candidate D

cos 36

sin 1 5

36tan

15

1·176 rad or 67 4

k a

k a

a

a

Candidate E

cos 15

sin 36

36tan

15

1·176 rad or 67 4

k a

k a

a

a

Candidate F

cos 36

sin 15

15tan

36

5·888 rad or 337 4

k a

k a

a

a

Response 3: Labelling incorrect, sin (A – B) = sin A cos B – cos A sin B from formula list.

Candidate G

sin A cos B – cos A sin B

cos 36

sin 1 5

15tan

36

0·39479 rad or 22 6

k k

k a

k a

a

a

Candidate H

sin A cos B – cos A sin B

cos1 5 36

sin1 5 1 5

15tan1 5

36

1 5 0·39479 rad or 22 6

k k

k t

k t

t

t

Candidate I

sin cos – cos sin

cos B 36

sin B 1 5

15tan B

36

B 0·39479 rad or 22 6

k A B k A B

k

k

Candidate J

1

39sin(1 5 0 395) 100

100sin(1 5 0 395)

39

1001 5 0 395 sin

39

t

t

t

Candidate K

1

39sin(1 5 0 395) 100

391 5 0 395 sin

100

t

t

[END OF MARKING INSTRUCTIONS]

1

×

2

4

1

×

2

×

4

1

×

2

4

1

1 1

2

4 ×

2

×

4

2

×

4

1

1

^

2

3

4

Does not satisfy

equations at 2

1

^

2

×

3

^

4 ×

1

2

3

×

4 ×

5×

6

7

×

8 ×

1

6

×

7

×

8 ×

1

2015 Mathematics Higher Finalised Marking Instructions · Page 2 General Comments These marking...

Documents

Transcript of 2015 Mathematics Higher Finalised Marking Instructions · Page 2 General Comments These marking...